BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 5.2, Problem 16E
To determine

To Calculate: The left Riemann sum Ln and right Riemann sums Rn for the integral 02ex2dx for n=5,10,50,and100 using the calculator.

To Find: The two numbers which the integral 02ex2dx lies between that two numbers.

To Explain: the statement for the integral 12ex2dx.

Expert Solution

Answer to Problem 16E

The value of right Riemann sums Rn for the integral 02ex2dx for n=5,10,50,and100 is tabulated in table 1.

The value of left Riemann sums Ln for the integral 02ex2dx for n=5,10,50,and100 is tabulated in table 2.

The value of the integral lies between 0.09976 and 0.09956.

No, similar statement cannot be made for the integral 12ex2dx, since it is aincreasing function on limit [1,0] and decreasing function onlimit [0,2].

Explanation of Solution

Given:

The integral function as 02ex2dx.

Number of rectangles n=5,10,50and100.

Calculation:

Show the Equation of the integral as follows:

02ex2dx (1)

Consider the value of the function f(x)=ex2 (2)

Calculate the value of the function f(x)=ex2 within limit [0,2].

Substitute 0 for x in Equation (6).

f(1)=e(0)2=1

Substitute 2 for x in Equation (6).

f(2)=e(2)2=e4=0.183

The function f(x)=ex2 is a decreasing function within limits [0,2].

For n=5

Find the width (Δx) using the relation:

Δx=ban (3)

Here, the upper limit is b, the lower limit is a, and the number of rectangles is n.

The limits of the integral 02ex2dx, a=0 and b=2.

Substitute 2 for b, 5 for n, and 0 for a in Equation (3).

Δx=205=25

Calculate the right Riemann Sum for n=5,10,50and100.

The right endpoints are x1=25, x2=45, x3=65, x4=85 and x5=2.

The expression to find the right Riemann sum Rn as shown below:

Rn=i=1nf(xi)Δx=f(x1)Δx+f(x2)Δx+...+f(xn)Δx (4)

Here, the right endpoint height of first rectangle is f(x1), the width is Δx, height of right endpoint of second rectangle is f(x2), and left endpoint height of nth rectangle is f(xn).

Calculate the value of f(x1), f(x2), f(x3), f(x4), and f(x5) using the Equation (2).

Substitute 25 for x1 in Equation (2).

f(25)=e(25)2=e425=0.8521

Substitute 45 for x2 in Equation (2).

f(45)=e(45)2=e1625=0.5272

Substitute 65 for x3 in Equation (2).

f(65)=e(65)2=e3625=0.2369

Substitute 85 for x4 in Equation (2).

f(85)=e(85)2=e6425=0.0773

Substitute 105 for x5 in Equation (2).

f(2)=e(105)2=e10025=0.0183

Calculate the right Riemann sum using calculator.

Substitute 25 for Δx, 5 for n, and values of right end points in Equation (4).

R5=i=15f(xi)Δx=f(x1)Δx+f(x2)Δx+...+f(x5)Δx=f(25)×25+f(45)×25+...+f(2)×25=[(0.8521×0.4)+(0.5272×0.4)+(0.2369×0.4)+(0.0773×0.4)+(0.0183×0.4)]

=0.68479

The value of right Riemann Sum for n=5 is 0.68479_.

Similarly calculate the right Riemann sum for n=10,50,100.

Tabulate the values of right Riemann sum for n=5,10,50and100 as shown in table 1.

nRn
50.68479
100.68115
500.19652
1000.09956

Table 1

Calculate the left Riemann Sum for n=5,10,50and100.

The left endpoints are x0=0, x1=25, x2=45, x3=65 and x4=85.

The expression to find the left Riemann sum Ln as shown below:

Ln=i=1nf(xi)Δx=f(x0)Δx+f(x1)Δx+...+f(xn1)Δx (5)

Here, the left endpoint height of first rectangle is f(x0), the width is Δx, height of left endpoint of second rectangle is f(x1), and left endpoint height of nth rectangle is f(xn1).

Calculate the value of f(x0), f(x1), f(x2), f(x3), and f(x4) using the Equation (2).

Substitute 0 for x0 in Equation (2).

f(0)=e(0)2=1

Substitute 25 for x1 in Equation (2).

f(25)=e(25)2=e425=0.8521

Substitute 45 for x2 in Equation (2).

f(45)=e(45)2=e1625=0.5272

Substitute 65 for x3 in Equation (2).

f(65)=e(65)2=e3625=0.2369

Substitute 85 for x4 in Equation (2).

f(85)=e(85)2=e6425=0.0773

Calculate the left Riemann sum using calculator.

Substitute 25 for Δx, 5 for n, and values of left end points in Equation (5).

L5=i=051f(xi)Δx=f(x0)Δx+f(x1)Δx+...+f(x4)Δx=f(0)×25+f(25)×25+f(45)×25+...+f(85)×25=[(1×0.4)+(0.8521×0.4)+(0.5272×0.4)+(0.2369×0.4)+(0.0773×0.4)]

L5=1.07746

The value of left Riemann Sum for n=5 is 1.07746_.

Similarly calculate the left Riemann sum for n=10,50,100.

Tabulate the values of left Riemann sum for n=5,10,50and100 as shown in table 2.

nLn
51.07746
100.80758
500.19809
1000.09976

Table 2

Refer to Table 1 and Table 2.

The value of the integral lies as shown below:

Ln>02ex2dx>Rn0.09976>02ex2dx>0.09956

The value of the integral lies between 0.09976 and 0.09956.

whether similar statement can be made for the integral 12ex2dx.

Consider the value of the function f(x)=ex2 (6)

Calculate the value of the function f(x)=ex2 within limit [1,2].

Substitute 1 for x in Equation (6).

f(1)=e(1)2=e1=0.3678

Substitute 0 for x in Equation (6).

f(0)=e(0)2=1

Substitute 2 for x in Equation (6).

f(2)=e(2)2=e4=0.183

No, similar statement cannot be made for the integral 12ex2dx, since the function f(x)=ex2 is a increasing function on limit [1,0] and decreasing function on limit [0,2].

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